In our tutorials about Electromagnetism, we saw that when an electric current flows through a conductor, a magnetic field is generated around itself.
We also saw that if this single wire conductor is moved or rotated within a stationary magnetic field, an “EMF”, (ElectroMotive Force) is induced within the conductor due to the movement of the conductor through the magnetic flux.
From this tutorial we learnt that a relationship exists between Electricity and Magnetism giving us, as Michael Faraday discovered the effect of “Electromagnetic Induction” and it is this basic principal that electrical machines and generators use to generate a Sinusoidal Waveform for our mains supply.
In the Electromagnetic Induction, tutorial we said that when a single wire conductor moves through a permanent magnetic field thereby cutting its lines of flux, an EMF is induced in it.
However, if the conductor moves in parallel with the magnetic field in the case of points A and B, no lines of flux are cut and no EMF is induced into the conductor, but if the conductor moves at right angles to the magnetic field as in the case of points C and D, the maximum amount of magnetic flux is cut producing the maximum amount of induced EMF.
Also, as the conductor cuts the magnetic field at different angles between points A and C, 0 and 90^{o} the amount of induced EMF will lie somewhere between this zero and maximum value. Then the amount of emf induced within a conductor depends on the angle between the conductor and the magnetic flux as well as the strength of the magnetic field.
An AC generator uses the principal of Faraday’s electromagnetic induction to convert a mechanical energy such as rotation, into electrical energy, a Sinusoidal Waveform. A simple generator consists of a pair of permanent magnets producing a fixed magnetic field between a north and a south pole. Inside this magnetic field is a single rectangular loop of wire that can be rotated around a fixed axis allowing it to cut the magnetic flux at various angles as shown below.
As the coil rotates anticlockwise around the central axis which is perpendicular to the magnetic field, the wire loop cuts the lines of magnetic force set up between the north and south poles at different angles as the loop rotates. The amount of induced EMF in the loop at any instant of time is proportional to the angle of rotation of the wire loop.
As this wire loop rotates, electrons in the wire flow in one direction around the loop. Now when the wire loop has rotated past the 180^{o} point and moves across the magnetic lines of force in the opposite direction, the electrons in the wire loop change and flow in the opposite direction. Then the direction of the electron movement determines the polarity of the induced voltage.
So we can see that when the loop or coil physically rotates one complete revolution, or 360^{o}, one full sinusoidal waveform is produced with one cycle of the waveform being produced for each revolution of the coil. As the coil rotates within the magnetic field, the electrical connections are made to the coil by means of carbon brushes and sliprings which are used to transfer the electrical current induced in the coil.
The amount of EMF induced into a coil cutting the magnetic lines of force is determined by the following three factors.
We know that the frequency of a supply is the number of times a cycle appears in one second and that frequency is measured in Hertz. As one cycle of induced emf is produced each full revolution of the coil through a magnetic field comprising of a north and south pole as shown above, if the coil rotates at a constant speed a constant number of cycles will be produced per second giving a constant frequency. So by increasing the speed of rotation of the coil the frequency will also be increased. Therefore, frequency is proportional to the speed of rotation, ( ƒ ∝ Ν ) where Ν = r.p.m.
Also, our simple single coil generator above only has two poles, one north and one south pole, giving just one pair of poles. If we add more magnetic poles to the generator above so that it now has four poles in total, two north and two south, then for each revolution of the coil two cycles will be produced for the same rotational speed. Therefore, frequency is proportional to the number of pairs of magnetic poles, ( ƒ ∝ P ) of the generator where P = is the number of “pairs of poles”.
Then from these two facts we can say that the frequency output from an AC generator is:
Where: Ν is the speed of rotation in r.p.m. P is the number of “pairs of poles” and 60 converts it into seconds.
The EMF induced in the coil at any instant of time depends upon the rate or speed at which the coil cuts the lines of magnetic flux between the poles and this is dependant upon the angle of rotation, Theta ( θ ) of the generating device. Because an AC waveform is constantly changing its value or amplitude, the waveform at any instant in time will have a different value from its next instant in time.
For example, the value at 1ms will be different to the value at 1.2ms and so on. These values are known generally as the Instantaneous Values, or V_{i} Then the instantaneous value of the waveform and also its direction will vary according to the position of the coil within the magnetic field as shown below.
The instantaneous values of a sinusoidal waveform is given as the “Instantaneous value = Maximum value x sin θ ” and this is generalized by the formula.
Where, V_{max} is the maximum voltage induced in the coil and θ = ωt, is the rotational angle of the coil with respect to time.
If we know the maximum or peak value of the waveform, by using the formula above the instantaneous values at various points along the waveform can be calculated. By plotting these values out onto graph paper, a sinusoidal waveform shape can be constructed.
In order to keep things simple we will plot the instantaneous values for the sinusoidal waveform at every 45^{o} of rotation giving us 8 points to plot. Again, to keep it simple we will assume a maximum voltage, V_{MAX} value of 100V. Plotting the instantaneous values at shorter intervals, for example at every 30^{o} (12 points) or 10^{o} (36 points) for example would result in a more accurate sinusoidal waveform construction.
Coil Angle ( θ )  0  45  90  135  180  225  270  315  360 
e = Vmax.sinθ  0  70.71  100  70.71  0  70.71  100  70.71  0 
The points on the sinusoidal waveform are obtained by projecting across from the various positions of rotation between 0^{o} and 360^{o} to the ordinate of the waveform that corresponds to the angle, θ and when the wire loop or coil rotates one complete revolution, or 360^{o}, one full waveform is produced.
From the plot of the sinusoidal waveform we can see that when θ is equal to 0^{o}, 180^{o} or 360^{o}, the generated EMF is zero as the coil cuts the minimum amount of lines of flux. But when θ is equal to 90^{o} and 270^{o} the generated EMF is at its maximum value as the maximum amount of flux is cut.
Therefore a sinusoidal waveform has a positive peak at 90^{o} and a negative peak at 270^{o}. Positions B, D, F and H generate a value of EMF corresponding to the formula e = Vmax.sinθ.
Then the waveform shape produced by our simple single loop generator is commonly referred to as a Sine Wave as it is said to be sinusoidal in its shape. This type of waveform is called a sine wave because it is based on the trigonometric sine function used in mathematics, ( x(t) = Amax.sinθ ).
When dealing with sine waves in the time domain and especially current related sine waves the unit of measurement used along the horizontal axis of the waveform can be either time, degrees or radians. In electrical engineering it is more common to use the Radian as the angular measurement of the angle along the horizontal axis rather than degrees. For example, ω = 100 rad/s, or 500 rad/s.
The Radian, (rad) is defined mathematically as a quadrant of a circle where the distance subtended on the circumference equals the radius (r) of the circle. Since the circumference of a circle is equal to 2π x radius, there must be 2π radians around a 360^{o} circle, so 1 radian = 360^{o}/2π = 57.3^{o}. In electrical engineering the use of radians is very common so it is important to remember the following formula.
Using radians as the unit of measurement for a sinusoidal waveform would give 2π radians for one full cycle of 360^{o}. Then half a sinusoidal waveform must be equal to 1π radians or just π (pi). Then knowing that pi, π is equal to 3.142 or 22÷7, the relationship between degrees and radians for a sinusoidal waveform is given as.
Applying these two equations to various points along the waveform gives us.
The conversion between degrees and radians for the more common equivalents used in sinusoidal analysis are given in the following table.
Degrees  Radians  Degrees  Radians  Degrees  Radians 
0^{o}  0  135^{o} 
3π
4

270^{o} 
3π
2

30^{o} 
π
6

150^{o} 
5π
6

300^{o} 
5π
3

45^{o} 
π
4

180^{o}  π  315^{o} 
7π
4

60^{o} 
π
3

210^{o} 
7π
6

330^{o} 
11π
6

90^{o} 
π
2

225^{o} 
5π
4

360^{o}  2π 
120^{o} 
2π
3

240^{o} 
4π
3

The velocity at which the generator rotates around its central axis determines the frequency of the sinusoidal waveform. As the frequency of the waveform is given as ƒ Hz or cycles per second, the waveform has angular frequency, ω, (Greek letter omega), in radians per second. Then the angular velocity of a sinusoidal waveform is given as.
and in the United Kingdom, the angular velocity or frequency of the mains supply is given as:
in the USA as their mains supply frequency is 60Hz it is therefore: 377 rad/s
So we now know that the velocity at which the generator rotates around its central axis determines the frequency of the sinusoidal waveform and which can also be called its angular velocity, ω. But we should by now also know that the time required to complete one revolution is equal to the periodic time, (T) of the sinusoidal waveform.
As frequency is inversely proportional to its time period, ƒ = 1/T we can therefore substitute the frequency quantity in the above equation for the equivalent periodic time quantity and substituting gives us.
The above equation states that for a smaller periodic time of the sinusoidal waveform, the greater must be the angular velocity of the waveform. Likewise in the equation above for the frequency quantity, the higher the frequency the higher the angular velocity.
A sinusoidal waveform is defined as: V_{m} = 169.8 sin(377t) volts. Calculate the RMS voltage of the waveform, its frequency and the instantaneous value of the voltage, (V_{i}) after a time of six milliseconds (6ms).
We know from above that the general expression given for a sinusoidal waveform is:
Then comparing this to our given expression for a sinusoidal waveform above of V_{m} = 169.8 sin(377t) will give us the peak voltage value of 169.8 volts for the waveform.
The waveforms RMS voltage is calculated as:
The angular velocity (ω) is given as 377 rad/s. Then 2πƒ = 377. So the frequency of the waveform is calculated as:
The instantaneous voltage V_{i} value after a time of 6mS is given as:
Note that the angular velocity at time t = 6mS is given in radians so we have to convert this into an equivalent angle in degrees and use this value instead to calculate the instantaneous voltage value. The angle in degrees is therefore given as:
Then the generalised format used for analysing and calculating the various values of a Sinusoidal Waveform is as follows:
In the next tutorial about Phase Difference we will look at the relationship between two sinusoidal waveforms that are of the same frequency but pass through the horizontal zero axis at different time intervals.
Very helpful artical for beginner s …thank you.
Hi there
Please advise how I calculate the maximum value of a wave if has 3ms and 60 volts?
Vp = 60 Vrms x 1.414 = 84.8 Vp its that simple, you should be able to do this yourself 🙁
A big thankyou! This is the first time after many years that I discovered the practical application of trigonometry.
You are the best math teacher I have known.
Once again thank you.
Thanks 🙂
thanks a lot my teacher ♥
Want to know about AC current
What is the value of sin in an equation?.
On one of the last questions for instantaneous values the sum is
Vi= 169.8 sin ( 377x6ms), how do you get 130 v as the answer?.
This is a fantastic website.
Thanks Jim.
When a coil rotates inside a magnetic field, as in the case of an alternator, the voltage varies sinusoidally with time producing an AC voltage. The coil will take a certain amount of time to rotate around a full 360 degrees. The value of the voltage at any instant of time as the coil rotates is called the instantaneous voltage.
The instantaneous voltage of the coil at any point in time around the full 360 degree circle is given as:
Vt = Vmax.sin(wt)
where:
Vt is the instantaneous voltage
Vmax is the maximum voltage the coil can produce
sin stands for the sine function
w (omega) is the angular frequency in radians of the coil defined as 2pif
t is time in seconds from zero radians
then the sine angle of (377 x 0.006) = 2.262 radians
2.262 radians = 129.59 degrees, as one radian equals 57.3 degrees
sine(129.59) = 0.771
169.8 x 0.771 = 130.8 volts
Sorry to be stupid, but at the bottom of your answer you say that sine(129.59)= 0.771.
How does it equal that?.
What is the value of sin?. And what does the wt stand for in theta?, does it mean degree for w and time for t?, if so how do I workout time?. These are for Vi calculations.
Thanks Jim.
I’m confused as this article http://www.physbot.co.uk/magneticfieldsandinduction.html states that when the conductor is parallel to the magnetic field the maximum amount of magnetic flux is cut. This is in contrast to what this article seems to be saying. Also, the drawing here shows the conductor’s points of A and B perpendicular to the magnetic filed, not parallel as is stated. Likewise, points C and D are said to be at a right angle with the magnetic field when they are clearly parallel.
What?, firstly I am not interested in commenting on other peoples articles, Secondly regarding the first image, points A and B run parallel, that is in the same direction as the lines of flux so the minimum amount of lines of flux are cut, therefore the minimum amount of voltage is induced. Points C and D as shown cut across the lines of flux, that is at 90 degrees, therefore the maximum number of lines of flux are cut, therefore the maximum amount of voltage is induced.
Thanks for the reply, Wayne. Perhaps I’m looking at this the wrong way. When looking at points C and D, I imagine them to be connected as they are two corners of the conductor. If that is the case how would they cut the lines of flux since they run parallel with them? Is there something I’m missing? Does the conductor produce some kind of force that would cut the lines of flux at 90 degrees? Or is it the conductor itself that disrupts the magnetic field? I’m pretty knew to this so hopefully you can make sense of my confusion.
Loops A+B and C+D are rotating within the magnetic field as indicated.
very good article… thanks 🙂
can u give me equation of 3d sinusoidal wave form………plz